Prove the railway metric is a metric: trouble proving d(x, y) = d(y, x)

Asked Oct 09 '19 at 10:16

Active Oct 09 '19 at 10:19

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I'm struggling to prove symmetry for the railway metric.

Let $d:\mathbb{R}^{2}\times\mathbb{R}^{2}\rightarrow[0,\infty)$

$$d(x,y)=\begin{cases}\|x-y||&\text{ if }x,y,0\text{ are collinear}.\\\|x−0||+\|0-y||&\text{ otherwise}.\end{cases}$$

I have managed to show $d(x,y)=\|x-y\|=\|x-x\|=\|0\|=0.$

edited Oct 09 '19 at 10:19

Floris Claassens

asked Oct 09 '19 at 10:16

$x,y,0$ are collinear iff $y,x,0$ are collinear. What is the problem you are facing with symmetry? – Kavi Rama Murthy Oct 09 '19 at 10:19
I'm not sure if I am overthinking it. Since they are collinear do we just say that d(x,y)=||x-y||=||y-x||=d(y,x) ? – Oct 09 '19 at 10:22
Yes, that is the proof of symmetry. – Kavi Rama Murthy Oct 09 '19 at 10:23
oh! thank you, I think its the thought of collinear that I was forgetting about – Oct 09 '19 at 10:24

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